Equilibrium Terms | OCR A Level Chemistry Revision Notes 2017

Mole Fractions & Partial Pressure

Mole fractions and partial pressures are a feature of K_p calculations
- Put simply, the mole fraction is the fraction of the total number of moles that each chemical in a reaction is responsible for
- Partial pressure is the part of the total pressure that each chemical in a reaction is responsible for
The partial pressure of a gas is the pressure it exerts in a mixture of gases if it occupied the container on its own
Partial pressure is given the symbol p, so for a gas X, it is written as pX
The total pressure is the sum of the partial pressures (this is known as Daltons' Law)

Partial Pressures, downloadable AS & A Level Chemistry revision notes

The relationship between partial pressures in a mixture of gases and the total pressure

To find the partial pressure of a gas, you need two pieces of information
- The total pressure in the container
- The mole fraction
The mathematical relationships are as follows

$Partial pressure and mole fraction formulae, downloadable AS & A Level Chemistry revision notes$

Partial pressure and mole fraction expressions

Equilibrium Quantities

When dealing with equilibrium calculations there are certain calculations that you will be expected to be able to perform
- Calculating concentrations
  - Concentration (mol dm^-3) = $\frac{number of moles}{volume ({dm}^{3})}$
- Calculating equilibrium quantities
- Calculating mole fractions and partial pressures

Worked example

Calculating concentrations

Ethanoic acid and ethanol react according to the following equation:

CH₃COOH (I) + C₂H₅OH (I) ⇌ CH₃COOC₂H₅ (I) + H₂O (I)

At equilibrium, 500 cm³ of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.

Calculate the concentration of each chemical at equilibrium.

Answer

- [CH₃COOH] $= \frac{0.235}{0.500} =$ 0.470 mol dm^-3
- [C₂H₅OH] $= \frac{0.035}{0.500} =$ 0.070 mol dm^-3
- [CH₃COOC₂H₅] $= \frac{0.182}{0.500} =$ 0.364 mol dm^-3
- [H₂O] $= \frac{0.182}{0.500} =$ 0.364 mol dm^-3

Calculating equilibrium quantities

Some questions give the initial and equilibrium concentrations of the reactants but not the products
An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked example

Ethyl ethanoate is hydrolysed by water:

CH₃COOC₂H₅(I) + H₂O(I) ⇌ CH₃COOH(I) + C₂H₅OH(I)

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1dm³. At equilibrium 0.0654 mol of water are present.

Use this data to calculate the equilibrium concentrations of each chemical.

Answer

Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Equilibria Calculating Kc of ethyl ethanoate table

Step 2: Calculate the concentrations of the reactants and products

- [CH₃COOC₂H₅] $= \frac{0.0654}{1.00} =$ 0.0654 mol dm^-3
- [H₂O] $= \frac{0.0654}{1.00} =$ 0.0654 mol dm^-3
- [CH₃COOH] $= \frac{0.0346}{1.00} =$ 0.0346 mol dm^-3
- [C₂H₅OH] $= \frac{0.0346}{1.00} =$ 0.0346 mol dm^-3

Calculating mole fractions and partial pressures

These are two of the fundamental calculations associated with K_p calculations

Worked example

Working out mole fractions

A sample of 0.25 mole of nitrogen and 0.75 mole of hydrogen were reacted together to form ammonia. The equilibrium amount of nitrogen was 0.16 mol.

N₂(g) + 3H₂ (g) ⇌ 2NH₃(g)

Calculate the mole fractions of nitrogen, hydrogen and ammonia.

Answer

Write out the equation and record the initial, the change and the equilibrium amounts:

WE Mole Fractions Answer, downloadable AS & A Level Chemistry revision notes

Examiner Tip

You can check you have the mole fractions correct by adding them up and making sure they come to 1: 0.195 + 0.585 + 0.220 = 1

Worked example

The total pressure for the reaction, described above, of nitrogen and hydrogen to form ammonia was 150 kPa.

Calculate the partial pressure of each gas.

Answer

- Partial pressure of N₂ (g) = 0.195 x 150 = 29.25 kPa
- Partial pressure of H₂ (g) = 0.585 x 150 = 87.75 kPa
- Partial pressure of NH₃ (g) = 0.220 x 150 = 33.0 kPa

Examiner Tip

You can check that your partial pressures are correct by adding them up:
- 29.25 + 87.75 + 33.0 = 150
They should add up to the total pressure - if they don't, then there is at least one calculation wrong somewhere!

Equilibrium Terms (OCR A Level Chemistry)

Revision Note

Mole Fractions & Partial Pressure

Equilibrium Quantities

Worked example

Calculating equilibrium quantities

Worked example

Calculating mole fractions and partial pressures

Worked example

Examiner Tip

Worked example

Examiner Tip

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Richard